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probability!!!

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by kiennguyen » Mon Oct 26, 2009 6:48 pm
try this one! OA later! please clarify your answer!

Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
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by capnx » Mon Oct 26, 2009 7:12 pm
1/6?
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Re: probability!!!

by uttam.albela » Mon Oct 26, 2009 9:54 pm
kiennguyen wrote:try this one! OA later! please clarify your answer!

Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3
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good job!

by kiennguyen » Tue Oct 27, 2009 4:59 am
OA is 1/3
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Re: probability!!!

by chipbmk » Tue Oct 27, 2009 4:34 pm
Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3[/quote]

Can you please explain the portion I bolded above.

Why do we need to add those two ways?
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Re: probability!!!

by uttam.albela » Tue Oct 27, 2009 7:52 pm
chipbmk wrote:Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3
Can you please explain the portion I bolded above.

Why do we need to add those two ways?[/quote]

When you are selecting a correct set of letter and envelope(Say A+a), there are 2 different ways in which letters BCD can be arranged in bcd envelopes with none of these 3 letters in correct envelope as shown below.

BCD BCD
cdb dbc

So 2 ways when letter A is in correct envelopes and all other 3 letters in wrong envelope.
Similarly, with B in correct envelope, you have 2 ways and so on.

So 2*4 = 8 ways to achieve stated event.

Hopefully, the explanation is clear now.
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